小黑喜迎国庆胡吃海塞，等待降温的leetcode之旅:1026. 节点与其祖先之间的最大差值（祝中老黑蓝港今日约妹马到成功）

本文主要是介绍小黑喜迎国庆胡吃海塞，等待降温的leetcode之旅:1026. 节点与其祖先之间的最大差值（祝中老黑蓝港今日约妹马到成功），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

小黑暴力求解

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:def maxAncestorDiff(self, root: Optional[TreeNode]) -> int:if not root:return -1def pre_node(root):q = [root]display = []while q:node = q.pop()display.append(node)if node.right:q.append(node.right)if node.left:q.append(node.left)return displaydisplay = pre_node(root)max_ = -1for node in display:dis = pre_node(node)for n in dis:if abs(node.val-n.val) > max_:max_ = abs(node.val-n.val)return max_

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先序遍历法（python）

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:def maxAncestorDiff(self, root: Optional[TreeNode]) -> int:self.dfs = 0def dfs(node,min_,max_):if not node:returnmin_ = min(min_,node.val)max_ = max(max_,node.val)if not (node.left or node.right):self.dfs = max(self.dfs,max_- min_)dfs(node.left,min_,max_)dfs(node.right,min_,max_)dfs(root,root.val,root.val)return self.dfs

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先序遍历法（java）

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {int ans = 0;public int maxAncestorDiff(TreeNode root) {dfs(root,root.val,root.val);return this.ans;}public void dfs(TreeNode node,int min,int max){if(node == null){return ;}min = (min > node.val) ? node.val:min;max = (max < node.val) ? node.val:max;if(node.left == null && node.right == null){this.ans = (this.ans < max - min) ? max - min:this.ans;}else{dfs(node.left,min,max);dfs(node.right,min,max);}}
}